package com.husd.leetcode.linkedlist;

/**
 *
 * 61. 旋转链表
 *
 * 给定一个链表，旋转链表，将链表每个节点向右移动 k 个位置，其中 k 是非负数。
 *
 * 示例 1:
 *
 * 输入: 1->2->3->4->5->NULL, k = 2
 * 输出: 4->5->1->2->3->NULL
 * 解释:
 * 向右旋转 1 步: 5->1->2->3->4->NULL
 * 向右旋转 2 步: 4->5->1->2->3->NULL
 * 示例 2:
 *
 * 输入: 0->1->2->NULL, k = 4
 * 输出: 2->0->1->NULL
 * 解释:
 * 向右旋转 1 步: 2->0->1->NULL
 * 向右旋转 2 步: 1->2->0->NULL
 * 向右旋转 3 步: 0->1->2->NULL
 * 向右旋转 4 步: 2->0->1->NULL
 *
 * 来源：力扣（LeetCode）
 * 链接：https://leetcode-cn.com/problems/rotate-list
 * 著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。
 */
public class RotateRight {

    //TODO 值得一看
    public ListNode rotateRight(ListNode head, int k) {
        // base cases
        if (head == null) return null;
        if (head.next == null) return head;

        // close the linked list into the ring
        ListNode old_tail = head;
        int n;
        for(n = 1; old_tail.next != null; n++)
            old_tail = old_tail.next;
        old_tail.next = head;

        // find new tail : (n - k % n - 1)th node
        // and new head : (n - k % n)th node
        ListNode new_tail = head;
        for (int i = 0; i < n - k % n - 1; i++)
            new_tail = new_tail.next;
        ListNode new_head = new_tail.next;

        // break the ring
        new_tail.next = null;

        return new_head;
    }

    //我自己做的
    public ListNode rotateRight1(ListNode head, int n) {
        if(n == 0) return head;
        if(head == null) return null;

        int c = 0;
        ListNode curr = head;
        while(curr != null) {
            c++;
            curr = curr.next;
        }
        if(n == c) return head;
        while(n > c) n = n - c;

        ListNode first = head;
        ListNode tail = null;
        ListNode after = null;
        ListNode pre = null;
        int step = 0;
        while(first != null && step < n) {
            if(first.next == null) {
                tail = first;
            }
            first = first.next;
            step++;
        }
        while(first != null) {
            if(first.next == null) {
                tail = first;
            }
            first = first.next;
            if(after == null) {
                after = head.next;
                pre = head;
            } else {
                after = after.next;
                pre = pre.next;
            }
        }
        //表示要删除的是第一个元素
        if(pre == null) {
            return head;
        } else {
            pre.next = null;
            tail.next = head;
        }
        return after;
    }
}
